ROTFLMBO!!!!! Gawd you are sure full of funny statements.
If a ratio results in a reduced ability to perform an operation or function
for a given input, the ratio is less efficient.
Lever anaology: the longer the handle side of a lever, the more efficiently
a person can do a job for a given input force.
And changing final drive ratios and adding underdriven transmission ratios
results in greaterr efficiency.
For someone that keeps putting me down you have very little grasp of
reality. Obviously you are set in the ways you were brainwashed into.
Yes, I'm sure correct statements are pretty funny to someone without a
clue about physics.
No, it isn't more efficient at all as you have to move the lever farther
to accomplish the same amount of work on the other end. The longer the
handle the less force is required, but more distance is required in
direct proportion to the reduced force. Since work is a force acting
through a distance, the end result is no change in efficiency at all.
Since you aren't able to show the math or physics as you several times
claimed you could, here's a really simple example using a lever which
you claim to understand. I won't complicate things with gears, but they
are just two levers touching each other, but that adds a little
complexity and you don't deal well with that.
Definition: Efficiency = work out/work in (the / means "divided by")
Here's a reference simple enough that you should be able to understand
Definition: Work = force * distance (the * means "multiplied by")
Again, here's a pretty simple reference (high school level):
Now we have a lever that is 10 feet long with the fulcrum in the middle.
Let's say we put a 100 lb. weight on one end. If we move the other
end of the lever down by 1 ft we will lift the weight 1 ft. The work
out is the work done on the weight, which is 100 ft-lbs. The work in is
the work done by whatever pushed down on the lever and this is also 100
ft-lbs since we had a force of 100 lbs acting through a distance of 1
ft. Thus the efficiency is 100/100 or 1.
Now shift the fulcrum such that it is 9 feet from the weight and 1 foot
from whatever or whoever is pushing on the lever. Now to lift the 100
lb weight, we will have to push down with a force of 900 lb. You keep
claiming that we have now lost efficiency since it now requires more
force on the lever than before. However, let's do the math. Again,
let's raise the weight 1 ft to keep the work out the same as before at
100 ft-lbs. How far does the lever have to be pushed down to raise the
weight 1 ft? Since the ratio is 1:9, we have to push the lever down
only 1/9th of a foot to raise the weight one foot. So how much work
does this require? It requires force * distance or 900 lbs * 1/9 ft
which equals ... drum roll please ... 100 ft-lbs. Now lets calculate
our efficiency again. It is equal to 100/100 or 1. Doesn't this sound
familiar? Let's see, why yes, it is the SAME as when the ratio of the
lever was 1:1. Imagine that, the lever ratio doesn't change the
efficiency at all. :-)
If you equate understanding basic physics and math with being
brainwashed, then I'm guilty as charged. OK, I demonstrated pretty
clearly above that changing the ratio of a lever has NO affect on the
efficiency of the lever. It changes the ratio of forces, but has ZERO
affect on the efficiency of work/power transfer.
Now it is your turn to show a clear example that supports your assertion
that changing a ratio does change the efficency.
Here's the answer for you all. You're simply using the
word "efficient" differently.
To the engineer it's simply a mathematical formula. ie; 12 volts
in and 11 volts out equals blah blah blah.
To the mechanic; it's more practical. ie; if you can't reach the
ignition key to crank the starter, then it's blah blah blah
There it is, now all of you sit down, have a beer or whatever, and
get back to helping others and each other fix your cars efficiently :)
Exactly, that is what we've been trying to show Budd. He claimed his
argument was based on physics, so I used the physics definition of
efficiency. He then started making up his own definitions, but
continued to claim they were based on "physics."
Maybe his definition of physics is also other-worldly. :-)
Have you ever heard of a guy by the name of Archimedes? Look up his
information on levers.
The same rules apply to gearing. An underdrive ration is like a prybar with
a long handle and a short span from pivot to load, a small force moves a
large mass. OTOH, an overdrive is like a short handle and a longer span from
pivot to load and requires much more force to move a small mass.
To make a OD work, the axle ratio has to be lowered and it doesn't quite
make up the losses in the drive train . . .it can't as that would violate
the laws of levers and physics.
"Give me a lever long enough and a place to stand and I shall move the
World." (Archimedes, if I remember the quote correctly)
Your argument isn't very convincing in that, in your lever examples, you
appear to be equating force and mechanical advantage with energy loss,
and that just isn't inherently the case (the same being true with gear
ratios, which I know you correctly related to levers). The energy loss
(heat generated) has nothing to do with the mechanical advantage (the
distance/force trade-offs) but has everything to do with the friction in
the transfer of motion - nothing to do with the lengths of the levers
(size of the gears), and everything to do with the reduction of friction
at the points of sliding, pivoting, etc.
No offense, but this is exactly the kind of technically incorrect
discussion that seems to crop up with those who consider themselves
technically savvy in a *hands-on* sense but who put down those with
technical (in a *theoretical* sense) expertise. In the same way that
you would correctly invoke Archimedes, I would say "Yes, but you also
need to consider the law of the conservation of energy which you seem to
be ignoring (in your lever examples)". You could move the world with
Archimede's lever, but not very rapidly. You could use an opposite
lever to move a pea and move it very rapidly. The energy put into the
system to accomplish both may be the same - one has a lot of mass moving
very slowly, the other has a tiny mass moving very rapidly - the energy
calculations are the same - energy applied minus friction losses energy out.
Now - one implementation of the gears or levers may be more efficient
than the other, but it's all in how the friction losses are managed, and
not in the mechanical advantage per-se.
There's a place for both "hands on" and "theroetical" - and results are
best when they co-exist. Real world follows the laws of physics - what
some here disdainfully call "theoretical"; but what many are
sarcastically or disdainfully calling the "theoretical" *is/are* the
real world laws of physics.
(To reply by e-mail, replace the last letter of the alphabet in my
address with the letter 'x')
Frictional losses are a given in ANY transmission design, as are heat
losses and parasitic losses. So, why mention the obvious that, as an
engineer, you should know about already?
In an overdrive ratio, do you have more or less usable torque on the
output shaft of the gearset? Energy ( torque in), divided ( OD,
remember?) by the ratio equals output torque ( minus parasitic losses,
of course!!!!) proportionally reduced according to the leverage . . er,
ratio of the gearset.
Sorry, it seems I stepped on a toe or two . . .
You assume they are ignored because they are unmetioned. Name a
transmission design that generates no friction, heat or parasitic
losses????? Now, since there is no such thing, these losses are common
and may be set asiode for a discussion of RATIOS AND THEIR EFFICIENCY.
True, I don't deny that and I didn't deny it. It was a metaphor.
Yaes, as in a trebuchet, for example ... nope, I didn't deny this
either, bith both of your exaples are fine evdence of the effectiveness
of RATIOS dependent, of course, on whether you want to lift a
locomotive or throw rocks.
Gear ratios parallel that law.
Agreed, but why are you adding given and common known factors into a
discussion on the effectiveness of RATIOS if not to cloud the topic to
save your butt?
Assume a 20% loss in either transmission to parasitical loss for sake
A 1:1 direct drive with a 100 ft.lb. input torque has 80
ft.lb.available from the output shaft out of a theoretical maximum of
A 1:2 overdrive with a 100 ft.lb. input torque has only 40 ft.lb. left
out of a theoretical maximum of 50 ft.lb. (This is still theoretical
because it doesn't allow for the increased torque loss due to poor
Now, tell us again which has the best torque output to run 75 mph with?
Math is "real life", Bill. 1+1=2 And as long as math is a true science
it can prove physics.
Budd - I'll say this in the least offensive way I know how (Matt said it
and I said it before): Torque and energy are different units - losing
torque thru levering or gearing is not in itself energy loss. Energy is
work x speed x time. In your lever and gear examples, what you lose in
torque you gain in speed - the product of the two is proportional to
energy and will be the same - except - yes - for some energy loss due to
friction, but that friction is not inherent in the lever or gear
multiplication or division factor. Think about this example: Situation
1 - The fulcrum point is on a sliding friction bushing. Situation 2 -
The fulcrum point is on a roller bearing of some sort. If you truly
think it thru, you will realize that there are friction losses in both
cases, but it is not directly related to the ratios. You will also have
sliding friction between the gear teeth - but its amount is not a given
relation to the torque multiplication factor - it can be varied all over
the place depending on wheter it is lubed or not, and if lubed, what
lube is used.
I will give you that, simplisitically speaking, if you have two meshed
gears for one multiplication factor, you will have half the friction
losses than if you have a series of two pairs of meshed gears to achieve
the same final ratio (the difference between the OD and direct drive
scenarios). But *that's* where the added losses are coming from that
you're talking about - not in the ratios themselves.
In a nutshell, as I and Matt have already pointed out, your fallacy in
your explanation is in trying to equate torque multiplication with
energy multiplication. You trade off torque and speed, and energy being
the product of the two (with time factored in), well - you do the math.
The same kind of arguments occur about torque and horsepower because
of the same failure on people's part to understand what they are
mathematically (and in reality).
(To reply by e-mail, replace the last letter of the alphabet in my
address with the letter 'x')
Hold it!!! Hold the presses!! You just contradicted yourself. First you say
there is no energy loss then there is . . you don't know squat.
What the fixation with friction losses?? Is it some thrill for you? How many
times must I agree that there is parasitic losses before you set them aside
and discuss ratios?
Reading comprhension isn't your strong point is it? Again, how many times
will I hacve to tell you that I agree with this BEFORE IT GETS THRU YOUR
THICK SKULL ( no, I am not a politically correct person, thank you)
I did the math and the net effect is a major measurable loss of efficiency.
No, the same arguments happen because overeducated idiots like yourself have
the misconception you know everything.
The only reason you think I'm in error is because you added parasitic losses
into a discussion ABOUT RATIOS.
So, you failed to discuss the topic.
No, I don't have to post anything. Efficiency: the ratio of input to
reslutant output. If your output isn't within the design parameter, it's
either more or less efficient than required. Overdrive ratios are
inheriently inefficient. the proof is the almost mandatory need to shift
down to a lower ratio to climb a hill UNLESS YOU'RE TRAVELING AT A SPEED FAR
ABOVE THE POSTED SPEED LIMIT.
The true fact is it wouldn't matter if I did. You'd just wave your wallpaper
around and find some fault with it, most likely, my high school education.
Right, you don't have to, but you also can't. The need to shift down
has nothing to do with efficiency, and has everything to do with torque
multiplication which is an entirely separate topic.
That is where you are wrong. You post one single reputable reference
that supports your claim, and I'll concede that I'm wrong and you are
right. I don't care what your education level is. I only care that you
are claiming physics support for an argument that isn't based on
physics. I wouldn't care if you had a Ph.D., I'd still point out the
fallacy of your argument.
This is the classic "I lost the logical argument and now need to bail
out" statement. No matter what I do you will just keep picking on me.
I haven't heard that since grade school.
Less torque, yes, but not less power. Efficiency is related to
energy/work/power, not force/torque.
I see the problem now. You are talking about engine efficiency, not
gear ratio "efficiency" whatever that is. Keeping the engine in an
efficient RPM range is a different discussion altogether, but, again, it
doesn't matter how you keep it there (whether OD tranny and numerically
higher rear end ratio or 1:1 transmission and a lower numerically rear
end ratio. I thought that was your original argument, but you've now
changed horses mid-stream. I'm still waiting to see how moving the
point of reduction from the rear axle to the transmission makes a
fundamental change in efficiency of power transmission.
Then how come cars running north of me at the Bonneville salt flats run
faster with low numerical axles and direct trans ratios than cars with
higher numericals and OD ( case in point: Summers Brother's Goldenrod. A
still standing land speed record with a car that could not pull in 4th gear,
that is OD).
No, I'm talking ratio efficiency, but you keep avoiding / clouding the
You'll never find it. Your head is buried in the sands of overeducation.
Because the OVERALL gear ratio wasn't correct for the power curve of the
engine. Ideally, you want the overall ratio to be such that the engine
hits its peak horsepower RPM at exactly the same time that the car hits
its maximum aerodynamic drag speed. If the car is geared too tall,
which this one obviously is, then the aero drag on the car is increasing
faster than the power curve of the engine and once they cross, the car
will no longer gain speed.
This could easily be fixed by using a numerically higher rear axle
ratio. It has nothing to do with the transmission having a ratio less
than 1:1, it has everything to do with the OVERALL reduction ratio of
the driveline being incorrectly matched to the engine for this
particular situation (top speed on the level).
That is because the topic simply doesn't exist. Find us even one
reference that talks about a gear ratio efficiency. Just one...
I won't find it because it doesn't exist. And that is the same reason
that you can't find it and post a reference to it.
ROTFLMBO!!!! Well, you just stuck your foot in it. According to the Summers
Brothers, the car was wind tunnel designed to handle 500 mph, then the
engineers determined the engine outputs and ratios needed. And they blew it.
The real life facts call you a liar. Besides, lowering the final drive
ratios would have resulted in an unusable first gear due to the slick salt
conditions, a slower acceleration and no speed record.
Like I said, and your attitude continues to prove it: it wouldn't matter if
I did, mr bigot engineer.
Btw, how about you explain how the marvelous inventions of the past 6000
years came about before your ilk contaminated the Earth?
You have yet to show a real life fact. That is why most cars are pushed
several hundred yards down the run by another vehicle. Have you ever
actually been to Bonneville?
I knew you couldn't, but it has been fun watching you try to weasel out
of the hole you've dug.
You mean the attitude that simply is asking you to provide even one fact
to back up your claim?
Since you've conceded above that you are wrong and unable to support
your claim, my work here is finished. However, Bill is usually more
persistent than me so he'll probably entertain your ignorance a little
Methinks you are making the common confusion between power and force
(torque). A lever multiplies or divides force (or torque), but it
doesn't multiply or divide work or power.
You are the one proposing a violation of the laws of physics. Look up
the definition of work and then apply it to both ends of your lever and
report back your results.
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